# Consider the following unbalanced equation: Ca_{3} (PO_{4})_{2}(s)+3H_{2}SO_{4}(aq) \Rightarrow 3CaSO_{4}(s)+2H_{3}PO_{4}(aq). What

Consider the following unbalanced equation: $C{a}_{3}{\left(P{O}_{4}\right)}_{2}\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒3CaS{O}_{4}\left(s\right)+2{H}_{3}P{O}_{4}\left(aq\right)$. What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid $\left(98\mathrm{%}{H}_{2}S{O}_{4}$ by mass)?
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Cassandra Ramirez
Balance the equation:
$C{a}_{3}{\left(P{O}_{4}\right)}_{2\left(s\right)}+{3}_{H2}S{O}_{4\left(aq\right)}⇒3CaS{O}_{4\left(s\right)}+{2}_{H3PO4\left(aq\right)}$
Solve for the masses of the reactants:
${H}_{2}S{O}_{4}$

$C{a}_{3}{\left(P{O}_{4}\right)}_{2}$

$=3.2molC{a}_{3}{\left(P{O}_{4}\right)}_{2}×\frac{3mol{H}_{2}S{O}_{4}}{1molC{a}_{3}{\left(P{O}_{4}\right)}_{2}}=9.6mol{H}_{2}S{O}_{4}$
Since $9.6mol<10mol$,
$C{a}_{3}{\left(P{O}_{4}\right)}_{2}$
is the limiting reactant.
Step 3
Finally, solve for the masses of the products:
$3.2molC{a}_{3}{\left(P{O}_{4}\right)}_{2}×\frac{3molCaS{O}_{4}}{1molC{a}_{3}{\left(P{O}_{4}\right)}_{2}}×\frac{135.15gCaS{O}_{4}}{1molCaS{O}_{4}}$
$=1.3×{10}^{3}gCaS{O}_{4}$
zurilomk4

First, balance the reaction.
$C{a}_{3}{\left(P{O}_{4}\right)}_{2}\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒3CaS{O}_{4}\left(s\right)+2{H}_{3}P{O}_{4}\left(aq\right)$
The molar mass of calcium phosphate is $40.08×3+\left(30.97+16×4\right)×2=310.18\frac{g}{m}ol$
Therefore 1.0kg of calcium phosphate $=1000g$, and dividing by 310.18g/mol gives 3.22 moles.
The molar mass of the sulfuric acid is $1.01×2+32.07+16×4=98.09\frac{g}{m}ol$
The 1.0kg mixture is 98% sulfuric acid, which is 0.98kg or 980g. Dividing by the 98.09g/mol gives 9.99 moles.
From the balanced reaction, 1 mol calcium phosphate reacts with 3 mol sulfuric acid.
Therefore, 3.22 mol calcium phosphate will react with 3(3.22)=9.66mol sulfuric acid.
There are 9.99 mol sulfuric acid, so it is the excess reactant. We use the limiting reactant of calcium phosphate for the calculations.
From the reaction, 1 mol calcium phosphate forms 3 mol calcium sulfate and 2 mol phosphoric acid.
$\left(3.22molC{a}_{3}{\left(P{O}_{4}\right)}_{2}\right)\left(3\right)=9.66molCaS{O}_{4}$
$\left(3.22molC{a}_{3}{\left(P{O}_{4}\right)}_{2}\right)\left(2\right)=6.44mol{H}_{3}P{O}_{4}$